I am wondering if the sum
$$S=\sum_{k=1}^{\infty}\frac{1}{k(3k-1)}$$
has an exact expression. And when I plugged it into Wolfram Alpha it spitted out:
$$S=\frac{1}{6}\Big(-\sqrt{3} π + 9 \ln(3)\Big)$$
I am wondering how is the answer obtained? Is there a simple way of not using math beyond second year university to arrive to that answer?
Tuesday 17 February 2015
calculus - Evaluating $sum_{k=1}^{infty}frac{1}{k(3k-1)}$
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